A beaker contains a 25 ml solution of an unknown monoprotic acid that reacts in a 1:1 stoichiometric ratio with naoh. titrate the solution with naoh to determine the concentration of the acid. perform a titration by setting the concentration of the naoh solution and adding it to the acid solution using the different add base buttons. the equivalence point of the titration is passed when the solution color changes. the unknown sample can be titrated multiple times by pressing the retitrate button and starting over. enter the concentration of the unknown acid solution.

Respuesta :

Given:

Volume of the unknown monoprotic acid (HA) = 25 ml

To determine:

The concentration of the acid HA

Explanation:

The titration reaction can be represented as-

HA + NaOH → Na⁺A⁻ + H₂O

As per stoichiometry: 1 mole of HA reacts with 1 mole of NaOH

At equivalence point-

moles of HA = moles of NaOH

For a known concentration and volume of added NaOH we have:

moles of NaOH = M(NaOH) * V(NaOH)

Thus, the concentration of the unknown 25 ml (0.025 L) of HA would be-

Molarity of HA = moles of HA/Vol of HA

Molarity of HA = M(NaOH)*V(NaOH)/0.025 L


Answer:

The concentration of monoprotic acid on titration against NaOH was found to be 0.36 M.

Explanation:

The reaction of monoprotic acid (HA) with NaOH will give the reaction equation:

[tex]\rm HA\;+\;NaOH\;\rightarrow\;NaA\;+H_2O[/tex]

This states that equal moles of NaOH and HA are required in the neutralization reaction.

For determining the concentration of the monoprotic acid, the HA is titrated against 0.2 M, 45 ml NaOH.

So, the concentration of the 25 ml HA, which is titrated will be;

[tex]\rm M_1\;V_1\;=\;M_2\;V_2[/tex]

[tex]\rm M_1\;and\;V_1[/tex] are molarity and volume of NaOH

[tex]\rm M_2\;and\;V_2[/tex] are molarity and volume of HA.

[tex]\rm 0.2\;\times\;45\;=\;M_2\;\times\;25[/tex]

[tex]\rm M_2[/tex]  = 0.36 M

The concentration of monoprotic acid titrated against 0.2 M of 45 ml NaOH was found to be 0.36 M.

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